If $α, β$ are the roots of the equation $a x^{2} + b x + c = 0$ then the equation whose roots are $α + \frac{1}{β}$ and $β + \frac{1}{α}$ is

$a c x^{2} + (a + b) c x + (a + c)^{2} = 0$

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$a b x^{2} + (a + c) b x + (a + c)^{2} = 0$

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$a c x^{2} + (a + c) b x + (a + c)^{2} = 0$

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$N o n e o f t h e s e$

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Solution

The correct option is C
$a c x^{2} + (a + c) b x + (a + c)^{2} = 0$

Here $α + β =$ $- \frac{b}{a}$ and $α β =$ $\frac{c}{a}$

If roots are $α + \frac{1}{β}, β + \frac{1}{α}$ , then sum of roots are

$= (α + \frac{1}{β}) + (β + \frac{1}{α}) = (α + β) + \frac{α + β}{α β} = - \frac{b}{a c} (a + c)$

and product $= (α + \frac{1}{β}) (β + \frac{1}{α})$

$= α β + 1 + 1 + \frac{1}{α β} = 2 + \frac{c}{a} + \frac{a}{c}$

$= \frac{2 a c + c^{2} + a^{2}}{a c} = \frac{(a + c)^{2}}{a c}$

Hence required equation is given by

$x^{2} + \frac{b}{a c} (a + c) x + \frac{(a + c)^{2}}{a c} = 0$

$\Rightarrow$ $a c x^{2} + (a + c) b x + (a + c)^{2} = 0$

Trick : Let a = 1, b = -3, c = 2,then $α$ = 1, $β$ = 2

$∴ α + \frac{1}{β} = \frac{3}{2} a n d β + \frac{1}{α} = 3$

$∴$ required equation must be

$(x - 3) (2 x - 3) = 0$ i.e. $2 x^{2} - 9 x + 9 = 0$

Here (a) gives this equation on putting

a = 1, b = -3, c = 2

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If $α, β$ are the roots of the equation $a x^{2} + b x + c = 0$ then the equation whose roots are $α + \frac{1}{β}$ and $β + \frac{1}{α}$ is

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